AP Inter 1st Year Maths Exercise 3b Solutions

Referring to the AP Inter 1st Year Maths Study Material Chapter Trigonometric Functions Exercise 3b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Trigonometric Functions Exercise 3b

I. Find the values of the trigonometric functions.

Question 1.
sin 765°
Solution:
sin 765° = sin (2 × 360° + 45°)
= sin 45°
= \(\frac{1}{\sqrt{2}}\) [∵ x ∈ Q₁]

Question 2.
cosec (- 1410°)
Solution:
cosec (- 1410°) = – cosec (1410°) [∵ x ∈ Q₄]
= – cosec (4 × 360° – 30°)
= – [- cosec 30°]
= cosec 30° = 2

Question 3.
tan \(\frac{19 \pi}{3}\)
Solution:
tan \(\frac{19 \pi}{3}\) = tan (6π + \(\frac{\pi}{3}\))
= tan \(\frac{\pi}{3}\) [∵ x ∈ Q₁] = \(\sqrt{3}\)

Question 4.
sin (- \(\frac{11 \pi}{3}\))
Solution:
sin (- \(\frac{11 \pi}{3}\)) = – sin (\(\frac{11 \pi}{3}\)) [∵ x ∈ Q₄]
= – sin (4π – \(\frac{\pi}{3}\))
= – [- sin \(\frac{\pi}{3}\)]
= sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)

Question 5.
cot (- \(\frac{15 \pi}{4}\))
Solution:
cot (- \(\frac{15 \pi}{4}\)) = – cot (\(\frac{15 \pi}{4}\)) [∵ x ∈ Q₄]
= – cot (4π – \(\frac{\pi}{4}\))
= – [- cot \(\frac{\pi}{4}\)]
= cot \(\frac{\pi}{4}\) = 1.

II. Find the values of other five trigonometric functions.

Question 1.
cos x = – ½, x lies in third quadrant.
Solution:
sin² x = 1 – cos² x
= 1 – (- \(\frac{1}{2}\))²
= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Therefore, sin x = – \(\frac{\sqrt{3}}{2}\) [∵ x ∈ Q₃]
cosec x ⇒ \(\frac{1}{\sin x}=-\frac{2}{\sqrt{3}}\),
sec x ⇒ \(\frac{1}{cos x}\)
= \(-\frac{2}{1}\) = – 2
tan x ⇒ \(\frac{\sin x}{\cos x}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\) = √3
cot x ⇒ \(\frac{\cos x}{\sin x}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}\)
= \(\frac{1}{\sqrt{3}}\)

Question 2.
sin x = \(\frac{3}{5}\), x lies in second quadrant.
Solution:
cos² x = 1 – sin² x
= 1 – (\(\frac{3}{5}\))²
= 1 – \(\frac{9}{25}\) = \(\frac{16}{25}\)
Therefore, cos x = – \(\frac{4}{5}\) [∵ x ∈ Q₂]

Question 3.
cot x = \(\frac{3}{4}\), x lies in third quadrant.
Solution:
cosec² x = 1 + cot² x

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Question 4.
sec x = \(\frac{13}{5}\), x lies in fourth quadrant.
Solution:
sec x = \(\frac{13}{5}\)
⇒ cos x = \(\frac{5}{13}\)

Question 5.
tan x = \(-\frac{5}{12}\), x lies in second quadrant.
Solution:
sec² x = 1 + tan² x